Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class KarenAndGame2 {
        public static boolean notMatch(int[][] g, int[][] cur) {
        for (int i = 0; i < g.length; ++i) {
            for (int j = 0; j < g[i].length; ++j) {
                if (g[i][j] != cur[i][j]) {
                    return true;
                }
            }
        }
        return false;
    }

    public static void main(String[] args) throws Exception {
        //ingresar in.readLine()
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        // para mostrar out.println(-1);
        PrintWriter out = new PrintWriter(System.out);
        //para separa por espacion st.nextToken()
        StringTokenizer st = new StringTokenizer(in.readLine());
        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());
        int[][] g = new int[n][m];
        for (int i = 0; i < n; ++i) {
            st = new StringTokenizer(in.readLine());
            for (int j = 0; j < m; ++j) {
                g[i][j] = Integer.parseInt(st.nextToken());
            }
        }
        StringBuilder sb1 = new StringBuilder();
        StringBuilder sb2 = new StringBuilder();
        int[][] cur1 = new int[n][m];
        int[][] cur2 = new int[n][m];
        int res1 = 0;
        int res2 = 0;
        for (int i = 0; i < n; ++i) {
            int minVal = g[i][0] - cur1[i][0];
            for (int j = 1; j < m; ++j) {
                if ((g[i][j] - cur1[i][j]) < minVal) {
                    minVal = (g[i][j] - cur1[i][j]);
                }
            }
            if (minVal > 0) {
                for (int j = 0; j < m; ++j) {
                    cur1[i][j] += minVal;
                }
                for (int j = 0; j < minVal; ++j) {
                    sb1.append("row " + (i + 1) + "\n");
                    res1++;
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            int minVal = g[0][i] - cur1[0][i];
            for (int j = 1; j < n; ++j) {
                if ((g[j][i] - cur1[j][i]) < minVal) {
                    minVal = (g[j][i] - cur1[j][i]);
                }
            }
            if (minVal > 0) {
                for (int j = 0; j < n; ++j) {
                    cur1[j][i] += minVal;
                }
                for (int j = 0; j < minVal; ++j) {
                    sb1.append("col " + (i + 1) + "\n");
                    res1++;
                }
            }
        }

        for (int i = 0; i < m; ++i) {
            int minVal = g[0][i] - cur2[0][i];
            for (int j = 1; j < n; ++j) {
                if ((g[j][i] - cur2[j][i]) < minVal) {
                    minVal = (g[j][i] - cur2[j][i]);
                }
            }
            if (minVal > 0) {
                for (int j = 0; j < n; ++j) {
                    cur2[j][i] += minVal;
                }
                for (int j = 0; j < minVal; ++j) {
                    sb2.append("col " + (i + 1) + "\n");
                    res2++;
                }
            }
        }
        for (int i = 0; i < n; ++i) {
            int minVal = g[i][0] - cur2[i][0];
            for (int j = 1; j < m; ++j) {
                if ((g[i][j] - cur2[i][j]) < minVal) {
                    minVal = (g[i][j] - cur2[i][j]);
                }
            }
            if (minVal > 0) {
                for (int j = 0; j < m; ++j) {
                    cur2[i][j] += minVal;
                }
                for (int j = 0; j < minVal; ++j) {
                    sb2.append("row " + (i + 1) + "\n");
                    res2++;
                }
            }
        }
        if (notMatch(g, cur1) && notMatch(g, cur2)) {
            out.println(-1);
        } else if (res1 < res2) {
            out.println(res1);
            out.println(sb1);
        } else {
            out.println(res2);
            out.println(sb2);
        }
        in.close();
        out.close();
    }

}